2018 Pascal Contest - CEMC · 2018 Pascal Contest Solutions Page 7 21. We determine the number of ways to get to each unshaded square in the grid obeying the given rules. In the rst

The CENTRE for EDUCATION

in MATHEMATICS and COMPUTINGcemc.uwaterloo.ca

2018 Pascal Contest(Grade 9)

Tuesday, February 27, 2018(in North America and South America)

Wednesday, February 28, 2018(outside of North America and South America)

Solutions

©2017 University of Waterloo

2018 Pascal Contest Solutions Page 2

1. When we arrange the five choices from smallest to largest, we obtain 1.2, 1.4, 1.5, 2.0, 2.1.Thus, 1.2 is the smallest.

Answer: (B)

2. Evaluating,2018− 18 + 20

2=

2000 + 20

2=

2020

2= 1010.

Answer: (A)

3. July 14 is 11 days after July 3 of the same year.Since there are 7 days in a week, then July 10 and July 3 occur on the same day of the week,namely Wednesday.July 14 is 4 days after July 10, and so is a Sunday.

Answer: (C)

4. Since the car is charged 3 times per week for 52 weeks, it is charged 3× 52 = 156 times.Since the cost per charge is $0.78, then the total cost is 156× $0.78 = $121.68.

Answer: (E)

5. Since3× 3× 5× 5× 7× 9 = 3× 3× 7× n× n

then

n× n =3× 3× 5× 5× 7× 9

3× 3× 7= 5× 5× 9 = 5× 5× 3× 3

Since n× n = 5× 5× 3× 3, then a possible value for n is n = 5× 3 = 15.

Answer: (A)

6. Solution 1Consider the 6×6 square as being made up of a 2×6 rectangle on the left and a 4×6 rectangleon the right.

2

6

4

6

Each of these rectangles is divided in half by its diagonal, and so is half shaded.Therefore, 50% of the total area is shaded.


Solution 2The entire 6× 6 square has area 62 = 36.Both shaded triangles have height 6 (the height of the square).The bases of the triangles have lengths 2 and 4:

2

6

4

6

The left-hand triangle has area 12× 2× 6 = 6.

The right-hand triangle has area 12× 4× 6 = 12.

Therefore, the total shaded area is 6 + 12 = 18, which is one-half (or 50%) of the area of theentire square.

Answer: (A)

7. There are 5 + 7 + 8 = 20 ties in the box, 8 of which are pink.When Stephen removes a tie at random, the probability of choosing a pink tie is 8

20which is

equivalent to 25.

Answer: (C)

8. The section of the number line between 0 and 5 has length 5− 0 = 5.Since this section is divided into 20 equal parts, the width of each part is 5

20= 1

4= 0.25.

Since S is 5 of these equal parts to the right of 0, then S = 0 + 5× 0.25 = 1.25.Since T is 5 of these equal parts to the left of 5, then T = 5− 5× 0.25 = 5− 1.25 = 3.75.Therefore, S + T = 1.25 + 3.75 = 5.

Answer: (E)

9. If ♥ = 1, then ∇ = ♥×♥×♥ = 1× 1× 1 = 1, which is not possible since ∇ and ♥ must bedifferent positive integers.If ♥ = 2, then ∇ = ♥×♥×♥ = 2× 2× 2 = 8, which is possible.If ♥ = 3, then ∇ = ♥×♥×♥ = 3× 3× 3 = 27, which is not possible since ∇ is less than 20.If ♥ is greater than 3, then ∇ will be greater than 27 and so ♥ cannot be greater than 3.Thus, ♥ = 2 and so ∇ = 8.This means that ∇×∇ = 8× 8 = 64.

Answer: (D)

10. The line that passes through (−2, 1) and (2, 5) has slope5− 1

2− (−2)=

4

4= 1.

This means that, for every 1 unit to the right, the line moves 1 unit up.Thus, moving 2 units to the right from the starting point (−2, 1) to x = 0 will give a rise of 2.Therefore, the line passes through (−2 + 2, 1 + 2) = (0, 3).

Answer: (C)


11. The complete central angle of a circle measures 360◦.This means that the angle in the circle graph associated with Playing is 360◦ − 130◦ − 110◦

or 120◦.

A central angle of 120◦ represents120◦

360◦=

1

3of the total central angle.

This means that the baby polar bear plays for1

3of the day, which is

1

3× 24 = 8 hours.

Answer: (C)

12. Of the given uniform numbers,

• 11 and 13 are prime numbers

• 16 is a perfect square

• 12, 14 and 16 are even

Since Karl’s and Liu’s numbers were prime numbers, then their numbers were 11 and 13 insome order.Since Glenda’s number was a perfect square, then her number was 16.Since Helga’s and Julia’s numbers were even, then their numbers were 12 and 14 in some order.(The number 16 is already taken.)Thus, Ioana’s number is the remaining number, which is 15.

Answer: (D)

13. Since the given equilateral triangle has side length 10, its perimeter is 3× 10 = 30.In terms of x, the perimeter of the given rectangle is x+ 2x+ x+ 2x = 6x.Since the two perimeters are equal, then 6x = 30 which means that x = 5.Since the rectangle is x by 2x, its area is x(2x) = 2x2.Since x = 5, its area is 2(52) = 50.

Answer: (B)

14. The average of the numbers 7, 9, 10, 11 is7 + 9 + 10 + 11

4=

37

4= 9.25, which is not equal to

18, which is the fifth number.

The average of the numbers 7, 9, 10, 18 is7 + 9 + 10 + 18

4=

44

4= 11, which is equal to 11, the

remaining fifth number.

We can check that the averages of the remaining three combinations of four numbers is notequal to the fifth number.Therefore, the answer is 11.

(We note that in fact the average of the original five numbers is7 + 9 + 10 + 11 + 18

5=

55

5= 11,

and when we remove a number that is the average of a set, the average does not change. Canyou see why?)

Answer: (D)

15. We would like to find the first time after 4:56 where the digits are consecutive digits in increasingorder.It would make sense to try 5:67, but this is not a valid time.Similarly, the time cannot start with 6, 7, 8 or 9.No time starting with 10 or 11 starts with consecutive increasing digits.Starting with 12, we obtain the time 12:34. This is the first such time.


We need to determine the length of time between 4:56 and 12:34.From 4:56 to 11:56 is 7 hours, or 7× 60 = 420 minutes.From 11:56 to 12:00 is 4 minutes.From 12:00 to 12:34 is 34 minutes.Therefore, from 4:56 to 12:34 is 420 + 4 + 34 = 458 minutes.

Answer: (A)

16. First, we note that we cannot have n ≤ 6, since the first 6 letters are X’s.After 6 X’s and 3 Y’s, there are twice as many X’s as Y’s. In this case, n = 6 + 3 = 9.After 6 X’s and 12 Y’s, there are twice as many Y’s as X’s. In this case, n = 6 + 12 = 18.The next letters are all Y’s (with 24 Y’s in total), so there are no additional values of n withn ≤ 6 + 24 = 30.At this point, there are 6 X’s and 24 Y’s.After 24 Y’s and 12 X’s (that is, 6 additional X’s), there are twice as many Y’s as X’s. In thiscase, n = 24 + 12 = 36.After 24 Y’s and 48 X’s (that is, 42 additional X’s), there are twice as many X’s as Y’s. In thiscase, n = 24 + 48 = 72.Since we are told that there are four values of n, then we have found them all, and their sumis 9 + 18 + 36 + 72 = 135.

Answer: (C)

17. We note that n = p2q2 = (pq)2.Since n < 1000, then (pq)2 < 1000 and so pq <

√1000 ≈ 31.6.

Finding the number of possible values of n is thus equivalent to finding the number of positiveintegers m with 1 ≤ m ≤ 31 <

√1000 that are the product of two prime numbers.

The prime numbers that are at most 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31.The distinct products of pairs of these that are at most 31 are:

2× 3 = 6 2× 5 = 10 2× 7 = 14 2× 11 = 22 2× 13 = 26

3× 5 = 15 3× 7 = 21

Any other product either duplicates one that we have counted already, or is larger than 31.Therefore, there are 7 such values of n.

Answer: (E)

18. Consider 4PQR.Since the sum of the angles in a triangle is 180◦, then

∠QPR + ∠QRP = 180◦ − ∠PQR = 180◦ − 120◦ = 60◦

Since ∠QPS = ∠RPS, then ∠RPS = 12∠QPR.

Since ∠QRS = ∠PRS, then ∠PRS = 12∠QRP .

Therefore,

∠RPS + ∠PRS = 12∠QPR + 1

2∠QRP

= 12(∠QPR + ∠QRP )

= 12× 60◦

= 30◦

Finally, ∠PSR = 180◦ − (∠RPS + ∠PRS) = 180◦ − 30◦ = 150◦.Answer: (E)


19. We recall that time =distance

speed. Travelling x km at 90 km/h takes

x

90hours.

Travelling x km at 120 km/h takesx

120hours.

We are told that the difference between these lengths of time is 16 minutes.

Since there are 60 minutes in an hour, then 16 minutes is equivalent to16

60hours.

Since the time at 120 km/h is 16 minutes less than the time at 90 km/h, thenx

90− x

120=

16

60.

Combining the fractions on the left side using a common denominator of 360 = 4×90 = 3×120,

we obtainx

90− x

120=

4x

360− 3x

360=

x

360.

Thus,x

360=

16

60.

Since 360 = 6× 60, then16

60=

16× 6

360=

96

360. Thus,

x

360=

96

360which means that x = 96.

Answer: (D)

20. We drop a perpendicular from R to X on PT , join R to T , and drop a perpendicular from Sto Y on RT .

P Q

R

S

T

8

2

13

13

YX

Since quadrilateral PQRX has three right angles (at P , Q and X), then it must have four rightangles and so is a rectangle. Its area is 8× 2 = 16.Next, 4RXT is right-angled at X.Since PQRX is a rectangle, then XR = PQ = 8 and PX = QR = 2.Since PX = 2, then XT = PT − PX = 8− 2 = 6.Thus, the area of 4RXT is 1

2×XT ×XR = 1

2× 6× 8 = 24.

By the Pythagorean Theorem, TR =√XT 2 +XR2 =

√62 + 82 =

√36 + 64 =

√100 = 10,

since TR > 0.Since4TSR is isosceles with ST = SR and SY is perpendicular to TR, then Y is the midpointof TR.Since TY = Y R = 1

2TR, then TY = Y R = 5.

By the Pythagorean Theorem, SY =√ST 2 − TY 2 =

√132 − 52 =

√169− 25 =

√144 = 12,

since SY > 0.Therefore, the area of 4STR is 1

2× TR× SY = 1

2× 10× 12 = 60.

Finally, the area of pentagon PQRST is the sum of the areas of the pieces, or 60+24+16 = 100.

Answer: (D)


21. We determine the number of ways to get to each unshaded square in the grid obeying the givenrules.In the first row, there is 1 way to get to each unshaded square: by starting at that square.In each row below the first, the number of ways to get to an unshaded square equals the sumof the number of ways to get to each of the unshaded squares diagonally up and to the leftand up and to the right from the given square. This is because any path passing through anyunshaded square needs to come from exactly one of these unshaded squares in the row above.In the second row, there are 2 ways to get to each unshaded square: 1 way from each of twosquares in the row above.In the third row, there are 2 ways to get to each of the outside unshaded squares and 4 waysto get to the middle unshaded square.Continuing in this way, we obtain the following number of ways to get to each unshaded square:

1 1 1

2 2

2 24

6 6

66 12

Since there are 6, 12 and 6 ways to get to the unshaded squares in the bottom row, then thereare 6 + 12 + 6 = 24 paths through the grid that obey the given rules.

Answer: (D)

22. Each wire has 2 ends.Thus, 13 788 wires have 13 788× 2 = 27 576 ends.In a Miniou circuit, there are 3 wires connected to each node.This means that 3 wire ends arrive at each node, and so there are 27 576÷ 3 = 9192 nodes.

Answer: (B)


23. The circle with centre P has radius 1 and passes through Q.This means that PQ = 1.

Therefore, the circle with diameter PQ has radius 12

and so has area π(12

)2= 1

4π.

To find the area of the shaded region, we calculate the area of the region common to bothcircles and subtract the area of the circle with diameter PQ.Suppose that the two circles intersect at X and Y .Join X to Y , P to Q, P to X, P to Y , Q to X, and Q to Y (Figure 1).By symmetry, the area of the shaded region on each side of XY will be the same.The area of the shaded region on the right side of XY equals the area of sector PXQY of theleft circle minus the area of 4PXY (Figure 2).

P Q

X

Y

T P Q

X

Y

T

Figure 1 Figure 2

Since each of the large circles has radius 1, then PQ = PX = PY = QX = QY = 1.This means that each of 4XPQ and 4Y PQ is equilateral, and so ∠XPQ = ∠Y PQ = 60◦.Therefore, ∠XPY = 120◦, which means that sector PXQY is 120◦

360◦= 1

3of the full circle, and

so has area 13π12 = 1

3π.

Lastly, consider 4PXY .Note that PX = PY = 1 and that ∠XPQ = ∠Y PQ = 60◦.Since 4PXY is isosceles and PQ bisects ∠XPY , then PQ is perpendicular to XY at T andXT = TY .By symmetry, PT = TQ. Since PQ = 1, then PT = 1

2.

By the Pythagorean Theorem in 4PTX (which is right-angled at T ),

XT =√PX2 − PT 2 =

√12 −

(12

)2=√

1− 14

=√

34

=√32

since XT > 0.Therefore, XY = 2XT =

√3.

The area of 4PXY equals 12(XY )(PT ) = 1

2(√

3)(12

)=√34

.

Now, we can calculate the area of the shaded region to the right of XY to be 13π −

√34

, thedifference between the area of sector PXQY and the area of 4PXT .Therefore, the area of the shaded region with the circle with diameter PQ removed is

2(

13π −

√34

)− 1

4π = 2

3π −

√32− 1

4π = 5

12π −

√32≈ 0.443

Of the given choices, this is closest to 0.44.

Answer: (E)


24. We refer to the students with height 1.60 m as “taller” students and to those with height 1.22 mas “shorter” students.For the average of four consecutive heights to be greater than 1.50 m, the sum of these fourheights must be greater than 4× 1.50 m = 6.00 m.If there are 2 taller and 2 shorter students, then the sum of their heights is 2×1.60 m+2×1.22 mor 5.64 m, which is not large enough.Therefore, there must be more taller and fewer shorter students in a given group of 4 consecu-tive students.If there are 3 taller students and 1 shorter student, then the sum of their heights is equal to3× 1.60 m + 1× 1.22 m or 6.02 m, which is large enough.Thus, in Mrs. Warner’s line-up, any group of 4 consecutive students must include at least 3taller students and at most 1 shorter student. (4 taller and 0 shorter students also give anaverage height that is greater than 1.50 m.)For the average of seven consecutive heights to be less than 1.50 m, the sum of these sevenheights must be less than 7× 1.50 m = 10.50 m.Note that 6 taller students and 1 shorter student have total height 6× 1.60 m + 1× 1.22 m or10.82 m and that 5 taller students and 2 shorter students have total height equal to5× 1.60 m + 2× 1.22 m or 10.44 m.Thus, in Mrs. Warner’s line-up, any group of 7 consecutive students must include at most 5taller students and at least 2 shorter students.Now, we determine the maximum possible length for a line-up. We use “T” to represent a tallerstudent and “S” to represent a shorter student.After some fiddling, we discover the line-up TTSTTTSTT.The line-up TTSTTTSTT has length 9 and has the property that each group of 4 consecutivestudents includes exactly 3 T’s and each group of 7 consecutive students includes exactly 5 Tsand so has the desired average height properties.We claim that this is the longest such line-up, which makes the final answer 9 or (D).Suppose that there was a line-up of length at least 10, and the first 10 heights in the line-upwere abcdefghjk. Consider the following table of heights:

a b c d e f gb c d e f g hc d e f g h jd e f g h j k

We explain why this table demonstrates that it is not possible to have at least 10 in the line-up.Each row in this table is a list of 7 consecutive people from the line-up abcdefghjk and so itssum is less than 10.50 m.Each column in this table is a list of 4 consecutive people from the line-up abcdefghjk and soits sum is greater than 6.00 m.The sum of the numbers in this table equals the sum of the sums of the 4 rows, which must beless than 4× 10.50 m or 42.00 m.The sum of the numbers in this table equals the sum of the sums of the 7 columns, which mustbe greater than 7× 6.00 m or 42.00 m.The sum cannot be both less than and greater than 42.00 m.This means that our assumption is incorrect, and so it must be impossible to have a line-up oflength 10 or greater.(There are a number of other ways to convince yourself that there cannot be more than 9students in the line-up.)

Answer: (D)


25. Suppose that m = 500 and 1 ≤ n ≤ 499 and 1 ≤ r ≤ 15 and 2 ≤ s ≤ 9 and t = 0.Since s > 0, then the algorithm says that t is the remainder when r is divided by s.Since t = 0, then r is a multiple of s. Thus, r = as for some positive integer a.Since r > 0, then the algorithm says that s is the remainder when n is divided by r.In other words, n = br + s for some positive integer b.But r = as, so n = bas+ s = (ba+ 1)s.In other words, n is a multiple of s, say n = cs for some positive integer c.Since n > 0, then r is the remainder when m is divided by n.In other words, m = dn+ r for some positive integer d.But r = as and n = cs so m = dcs+ as = (dc+ a)s.In other words, m is a multiple of s, say m = es for some positive integer e.But m = 500 and 2 ≤ s ≤ 9.Since m is a multiple of s, then s is a divisor of 500 and so the possible values of s are s = 2, 4, 5.(None of 1, 3, 6, 7, 8, 9 is a divisor of 500.)We know that r is a multiple of s, that r > s (because s is the remainder when n is dividedby r), and that 1 ≤ r ≤ 15.If s = 5, then r = 10 or r = 15.If s = 4, then r = 8 or r = 12.If s = 2, then r = 4, 6, 8, 10, 12, 14.

Suppose that s = 5 and r = 10.Since m = dn+ r, then 500 = dn+ 10 and so dn = 490.Therefore, n is a divisor of 490, is a multiple of 5 (because n = cs), must be greater thanr = 10, and must be 5 more than a multiple of 10 (because the remainder when n is dividedby r is s).Since 490 = 5× 2× 72, then the divisors of 490 that are multiples of 5 are 5, 10, 35, 70, 245, 490(these are 5 times the divisors of 2 × 72). Among these, those greater than r = 10 havingremainder 5 when divided by 10 are 35 and 245, and so the possible values of n in this case are35 and 245.

For each possible pair s and r, we determine the values of n that satisfy the following conditions:

• n is a divsior of 500− r,• n is a multiple of s,

• n is greater than r, and

• the remainder when n is divided by r is s.

We make a table:

s r 500− r Divisors of 500− r Possible nthat are multiples of s

5 10 490 = 5× 2× 72 5, 10, 35, 70, 245, 490 35, 2455 15 485 = 5× 97 5, 485 4854 8 492 = 4× 3× 41 4, 12, 164, 492 12, 164, 4924 12 488 = 4× 2× 61 4, 8, 244, 488 2442 4 496 = 2× 23 × 31 2, 4, 8, 16, 62, 124, 248, 496 622 6 494 = 2× 13× 19 2, 26, 38, 494 26, 38, 4942 8 492 = 2× 2× 3× 41 2, 4, 6, 12, 82, 164, 246, 492 822 10 490 = 2× 5× 72 2, 10, 14, 70, 98, 490 None2 12 488 = 2× 22 × 61 2, 4, 8, 122, 244, 488 1222 14 486 = 2× 35 2, 6, 18, 54, 162, 486 None


In each row, we calculate the prime factorization in 500−r in the third column, list the divisorsof 500−r that are multiples of s in the fourth column, and determine which of these are greaterthan r and give a remainder of s when divided by r in the fifth column.

Therefore, the possible values of n are 35, 245, 485, 12, 164, 492, 244, 62, 26, 38, 494, 82, 122, ofwhich there are 13.

Answer: (E)

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2018 Pascal Contest - CEMC · 2018 Pascal Contest Solutions Page 7 21. We determine the number of ways to get to each unshaded square in the grid obeying the given rules. In the rst